3.98 \(\int \frac {\sinh ^3(x)}{a+b \coth (x)} \, dx\)
Optimal. Leaf size=134 \[ -\frac {b \sinh ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a \cosh ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a b^2 \cosh (x)}{\left (a^2-b^2\right )^2}-\frac {a \cosh (x)}{a^2-b^2}-\frac {b^4 \tanh ^{-1}\left (\frac {\sinh (x) (a \coth (x)+b)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {b^3 \sinh (x)}{\left (a^2-b^2\right )^2} \]
[Out]
-b^4*arctanh((b+a*coth(x))*sinh(x)/(a^2-b^2)^(1/2))/(a^2-b^2)^(5/2)+a*b^2*cosh(x)/(a^2-b^2)^2-a*cosh(x)/(a^2-b
^2)+1/3*a*cosh(x)^3/(a^2-b^2)-b^3*sinh(x)/(a^2-b^2)^2-1/3*b*sinh(x)^3/(a^2-b^2)
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Rubi [A] time = 0.24, antiderivative size = 134, normalized size of antiderivative = 1.00,
number of steps used = 9, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used =
{3511, 3486, 2633, 2638, 3509, 206} \[ -\frac {b \sinh ^3(x)}{3 \left (a^2-b^2\right )}-\frac {b^3 \sinh (x)}{\left (a^2-b^2\right )^2}+\frac {a \cosh ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a b^2 \cosh (x)}{\left (a^2-b^2\right )^2}-\frac {a \cosh (x)}{a^2-b^2}-\frac {b^4 \tanh ^{-1}\left (\frac {\sinh (x) (a \coth (x)+b)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}} \]
Antiderivative was successfully verified.
[In]
Int[Sinh[x]^3/(a + b*Coth[x]),x]
[Out]
-((b^4*ArcTanh[((b + a*Coth[x])*Sinh[x])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2)) + (a*b^2*Cosh[x])/(a^2 - b^2)^2
- (a*Cosh[x])/(a^2 - b^2) + (a*Cosh[x]^3)/(3*(a^2 - b^2)) - (b^3*Sinh[x])/(a^2 - b^2)^2 - (b*Sinh[x]^3)/(3*(a^
2 - b^2))
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 2633
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
Rule 2638
Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]
Rule 3486
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])
Rule 3509
Int[sec[(e_.) + (f_.)*(x_)]/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Dist[f^(-1), Subst[Int[1/(a^
2 + b^2 - x^2), x], x, (b - a*Tan[e + f*x])/Sec[e + f*x]], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 + b^2, 0]
Rule 3511
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^
2), Int[(d*Sec[e + f*x])^m*(a - b*Tan[e + f*x]), x], x] + Dist[b^2/(d^2*(a^2 + b^2)), Int[(d*Sec[e + f*x])^(m
+ 2)/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 + b^2, 0] && ILtQ[m, 0]
Rubi steps
\begin {align*} \int \frac {\sinh ^3(x)}{a+b \coth (x)} \, dx &=\frac {\int (a-b \coth (x)) \sinh ^3(x) \, dx}{a^2-b^2}+\frac {b^2 \int \frac {\sinh (x)}{a+b \coth (x)} \, dx}{a^2-b^2}\\ &=-\frac {b \sinh ^3(x)}{3 \left (a^2-b^2\right )}+\frac {b^2 \int (a-b \coth (x)) \sinh (x) \, dx}{\left (a^2-b^2\right )^2}+\frac {b^4 \int \frac {\text {csch}(x)}{a+b \coth (x)} \, dx}{\left (a^2-b^2\right )^2}+\frac {a \int \sinh ^3(x) \, dx}{a^2-b^2}\\ &=-\frac {b^3 \sinh (x)}{\left (a^2-b^2\right )^2}-\frac {b \sinh ^3(x)}{3 \left (a^2-b^2\right )}+\frac {\left (a b^2\right ) \int \sinh (x) \, dx}{\left (a^2-b^2\right )^2}-\frac {b^4 \operatorname {Subst}\left (\int \frac {1}{a^2-b^2-x^2} \, dx,x,i (-i b-i a \coth (x)) \sinh (x)\right )}{\left (a^2-b^2\right )^2}-\frac {a \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cosh (x)\right )}{a^2-b^2}\\ &=-\frac {b^4 \tanh ^{-1}\left (\frac {(b+a \coth (x)) \sinh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {a b^2 \cosh (x)}{\left (a^2-b^2\right )^2}-\frac {a \cosh (x)}{a^2-b^2}+\frac {a \cosh ^3(x)}{3 \left (a^2-b^2\right )}-\frac {b^3 \sinh (x)}{\left (a^2-b^2\right )^2}-\frac {b \sinh ^3(x)}{3 \left (a^2-b^2\right )}\\ \end {align*}
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Mathematica [A] time = 0.83, size = 171, normalized size = 1.28 \[ \frac {3 b \sqrt {b-a} \left (a^3+a^2 b-5 a b^2-5 b^3\right ) \sinh (x)-3 a \sqrt {b-a} \left (3 a^3+3 a^2 b-7 a b^2-7 b^3\right ) \cosh (x)+24 b^4 \sqrt {a+b} \tan ^{-1}\left (\frac {a+b \tanh \left (\frac {x}{2}\right )}{\sqrt {b-a} \sqrt {a+b}}\right )+b (b-a)^{3/2} (a+b)^2 \sinh (3 x)-a (b-a)^{3/2} (a+b)^2 \cosh (3 x)}{12 (b-a)^{5/2} (a+b)^3} \]
Antiderivative was successfully verified.
[In]
Integrate[Sinh[x]^3/(a + b*Coth[x]),x]
[Out]
(24*b^4*Sqrt[a + b]*ArcTan[(a + b*Tanh[x/2])/(Sqrt[-a + b]*Sqrt[a + b])] - 3*a*Sqrt[-a + b]*(3*a^3 + 3*a^2*b -
7*a*b^2 - 7*b^3)*Cosh[x] - a*(-a + b)^(3/2)*(a + b)^2*Cosh[3*x] + 3*b*Sqrt[-a + b]*(a^3 + a^2*b - 5*a*b^2 - 5
*b^3)*Sinh[x] + b*(-a + b)^(3/2)*(a + b)^2*Sinh[3*x])/(12*(-a + b)^(5/2)*(a + b)^3)
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fricas [B] time = 0.46, size = 1859, normalized size = 13.87 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(x)^3/(a+b*coth(x)),x, algorithm="fricas")
[Out]
[1/24*((a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^6 + 6*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3
+ a*b^4 - b^5)*cosh(x)*sinh(x)^5 + (a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*sinh(x)^6 + a^5 + a^4*b
- 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5 - 3*(3*a^5 - a^4*b - 10*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 - 5*b^5)*cosh(x)^
4 - 3*(3*a^5 - a^4*b - 10*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 - 5*b^5 - 5*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b
^4 - b^5)*cosh(x)^2)*sinh(x)^4 + 4*(5*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^3 - 3*(3*a^5
- a^4*b - 10*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 - 5*b^5)*cosh(x))*sinh(x)^3 - 3*(3*a^5 + a^4*b - 10*a^3*b^2 - 6*a^
2*b^3 + 7*a*b^4 + 5*b^5)*cosh(x)^2 - 3*(3*a^5 + a^4*b - 10*a^3*b^2 - 6*a^2*b^3 + 7*a*b^4 + 5*b^5 - 5*(a^5 - a^
4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^4 + 6*(3*a^5 - a^4*b - 10*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 - 5
*b^5)*cosh(x)^2)*sinh(x)^2 + 24*(b^4*cosh(x)^3 + 3*b^4*cosh(x)^2*sinh(x) + 3*b^4*cosh(x)*sinh(x)^2 + b^4*sinh(
x)^3)*sqrt(a^2 - b^2)*log(((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 - 2*sqrt(a^2 - b^
2)*(cosh(x) + sinh(x)) + a - b)/((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 - a + b)) +
6*((a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^5 - 2*(3*a^5 - a^4*b - 10*a^3*b^2 + 6*a^2*b^3
+ 7*a*b^4 - 5*b^5)*cosh(x)^3 - (3*a^5 + a^4*b - 10*a^3*b^2 - 6*a^2*b^3 + 7*a*b^4 + 5*b^5)*cosh(x))*sinh(x))/((
a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^3 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^2*sinh(x) + 3*(a^
6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)*sinh(x)^2 + (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sinh(x)^3), 1/24*((a^
5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^6 + 6*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 -
b^5)*cosh(x)*sinh(x)^5 + (a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*sinh(x)^6 + a^5 + a^4*b - 2*a^3*b
^2 - 2*a^2*b^3 + a*b^4 + b^5 - 3*(3*a^5 - a^4*b - 10*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 - 5*b^5)*cosh(x)^4 - 3*(3*a
^5 - a^4*b - 10*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 - 5*b^5 - 5*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*
cosh(x)^2)*sinh(x)^4 + 4*(5*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^3 - 3*(3*a^5 - a^4*b -
10*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 - 5*b^5)*cosh(x))*sinh(x)^3 - 3*(3*a^5 + a^4*b - 10*a^3*b^2 - 6*a^2*b^3 + 7*
a*b^4 + 5*b^5)*cosh(x)^2 - 3*(3*a^5 + a^4*b - 10*a^3*b^2 - 6*a^2*b^3 + 7*a*b^4 + 5*b^5 - 5*(a^5 - a^4*b - 2*a^
3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^4 + 6*(3*a^5 - a^4*b - 10*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 - 5*b^5)*cosh
(x)^2)*sinh(x)^2 + 48*(b^4*cosh(x)^3 + 3*b^4*cosh(x)^2*sinh(x) + 3*b^4*cosh(x)*sinh(x)^2 + b^4*sinh(x)^3)*sqrt
(-a^2 + b^2)*arctan(sqrt(-a^2 + b^2)/((a + b)*cosh(x) + (a + b)*sinh(x))) + 6*((a^5 - a^4*b - 2*a^3*b^2 + 2*a^
2*b^3 + a*b^4 - b^5)*cosh(x)^5 - 2*(3*a^5 - a^4*b - 10*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 - 5*b^5)*cosh(x)^3 - (3*a
^5 + a^4*b - 10*a^3*b^2 - 6*a^2*b^3 + 7*a*b^4 + 5*b^5)*cosh(x))*sinh(x))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*
cosh(x)^3 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^2*sinh(x) + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*co
sh(x)*sinh(x)^2 + (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sinh(x)^3)]
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giac [A] time = 0.13, size = 163, normalized size = 1.22 \[ -\frac {2 \, b^{4} \arctan \left (-\frac {a e^{x} + b e^{x}}{\sqrt {-a^{2} + b^{2}}}\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {{\left (9 \, a e^{\left (2 \, x\right )} - 15 \, b e^{\left (2 \, x\right )} - a + b\right )} e^{\left (-3 \, x\right )}}{24 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac {a^{2} e^{\left (3 \, x\right )} + 2 \, a b e^{\left (3 \, x\right )} + b^{2} e^{\left (3 \, x\right )} - 9 \, a^{2} e^{x} - 24 \, a b e^{x} - 15 \, b^{2} e^{x}}{24 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(x)^3/(a+b*coth(x)),x, algorithm="giac")
[Out]
-2*b^4*arctan(-(a*e^x + b*e^x)/sqrt(-a^2 + b^2))/((a^4 - 2*a^2*b^2 + b^4)*sqrt(-a^2 + b^2)) - 1/24*(9*a*e^(2*x
) - 15*b*e^(2*x) - a + b)*e^(-3*x)/(a^2 - 2*a*b + b^2) + 1/24*(a^2*e^(3*x) + 2*a*b*e^(3*x) + b^2*e^(3*x) - 9*a
^2*e^x - 24*a*b*e^x - 15*b^2*e^x)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3)
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maple [A] time = 0.13, size = 197, normalized size = 1.47 \[ \frac {2 b^{4} \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {-a^{2}+b^{2}}}-\frac {32}{3 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3} \left (32 a +32 b \right )}-\frac {16}{\left (32 a +32 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {a}{2 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {b}{\left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {16}{\left (32 a -32 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {32}{3 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3} \left (32 a -32 b \right )}-\frac {a}{2 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {b}{\left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sinh(x)^3/(a+b*coth(x)),x)
[Out]
2*b^4/(a-b)^2/(a+b)^2/(-a^2+b^2)^(1/2)*arctan(1/2*(2*tanh(1/2*x)*b+2*a)/(-a^2+b^2)^(1/2))-32/3/(tanh(1/2*x)-1)
^3/(32*a+32*b)-16/(32*a+32*b)/(tanh(1/2*x)-1)^2+1/2/(a+b)^2/(tanh(1/2*x)-1)*a+1/(a+b)^2/(tanh(1/2*x)-1)*b-16/(
32*a-32*b)/(tanh(1/2*x)+1)^2+32/3/(tanh(1/2*x)+1)^3/(32*a-32*b)-1/2/(a-b)^2/(tanh(1/2*x)+1)*a+1/(a-b)^2/(tanh(
1/2*x)+1)*b
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(x)^3/(a+b*coth(x)),x, algorithm="maxima")
[Out]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?
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mupad [B] time = 1.86, size = 172, normalized size = 1.28 \[ \frac {{\mathrm {e}}^{-3\,x}}{24\,a-24\,b}+\frac {{\mathrm {e}}^{3\,x}}{24\,a+24\,b}-\frac {{\mathrm {e}}^{-x}\,\left (3\,a-5\,b\right )}{8\,{\left (a-b\right )}^2}-\frac {{\mathrm {e}}^x\,\left (3\,a+5\,b\right )}{8\,{\left (a+b\right )}^2}-\frac {b^4\,\ln \left (2\,a^3\,b-2\,a\,b^3+a^4-b^4+{\mathrm {e}}^x\,{\left (a+b\right )}^{7/2}\,\sqrt {a-b}\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}+\frac {b^4\,\ln \left (2\,a\,b^3-2\,a^3\,b-a^4+b^4+{\mathrm {e}}^x\,{\left (a+b\right )}^{7/2}\,\sqrt {a-b}\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sinh(x)^3/(a + b*coth(x)),x)
[Out]
exp(-3*x)/(24*a - 24*b) + exp(3*x)/(24*a + 24*b) - (exp(-x)*(3*a - 5*b))/(8*(a - b)^2) - (exp(x)*(3*a + 5*b))/
(8*(a + b)^2) - (b^4*log(2*a^3*b - 2*a*b^3 + a^4 - b^4 + exp(x)*(a + b)^(7/2)*(a - b)^(1/2)))/((a + b)^(5/2)*(
a - b)^(5/2)) + (b^4*log(2*a*b^3 - 2*a^3*b - a^4 + b^4 + exp(x)*(a + b)^(7/2)*(a - b)^(1/2)))/((a + b)^(5/2)*(
a - b)^(5/2))
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{3}{\relax (x )}}{a + b \coth {\relax (x )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(x)**3/(a+b*coth(x)),x)
[Out]
Integral(sinh(x)**3/(a + b*coth(x)), x)
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